Question

Upload Sailboat data and answer the questions below.

I) For an additional 3 feet increase how much heavier the sailboat gets? (10 points)

• a. 23.6 pounds
• b. 23600 pounds
• c. 3.03 pounds
• d. 3039 pounds
• e. 1.01 thousand pounds

II) After performing the Regression analysis on Sailboat data you are asked to pick one number that would best answer the question: Are these two variables length (in feet) and weight (in pounds) related or not? What is this number and why? (10 points)

• a. Coefficient 1.01; it tells us that for each foot the boat gains 1010 pounds, thus the variables are related.
• b. Intercept -18.0; tells us that the length and the weight is always related;
• c. Standard error is 4.4; and if we have an error of our prediction, there must be a relation between the variables.
• d. R-square; it is close to 1 and variables are most likely related
• e. P-Value; it is very small, thus the variables are most likely related.

III) Given a boat that is 30 feet in length use regression analysis and all available information in the table to predict its weight. Choose the interval prediction that can be deduced from the regression table. (10 points)

• a. 20000-25000
• b. 20000-27000
• c. 12000-15000
• d. 11000-13000
• e. 8000-17000

SAILBOAT DATA

Feet    Pounds 000
54      45
49      31
52      39
48      22
45      27
29      11
25      8
43      23
26      12
50      37
53      38
43      26
42      25
25      8
43      24
28      8
36      14
47      20
39      22
26      13

Answer 1

1)

Using R, we ran the linear regression on the given data.

> m = lm(Pounds ~ Feet, data = data)
> m

Call:
lm(formula = Pounds ~ Feet, data = data)

Coefficients:
(Intercept) Feet
-18.016 1.013

The slope coefficient of the feet is 1.013.

For an additional 3 feet increase, increase in weights = 3 * 1.013 = 3.039

• d. 3039 pounds

II)

The summary of the model is,

> summary(m)

Call:
lm(formula = Pounds ~ Feet, data = data)

Residuals:
Min 1Q Median 3Q Max
-9.5881 -1.7384 0.4698 2.6715 8.3219

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -18.01642 4.10059 -4.394 0.00035 ***
Feet 1.01286 0.09913 10.218 6.41e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.414 on 18 degrees of freedom
Multiple R-squared: 0.8529,   Adjusted R-squared: 0.8448
F-statistic: 104.4 on 1 and 18 DF, p-value: 6.406e-09

We see the p-value for the significance of the model is less than significance level of 0.05. This tells that the model is significant and variables Feet and Pound are significantly related.

• e. P-Value; it is very small, thus the variables are most likely related.

III)

The estimated regression line is,

Pound = (-18.016 + 1.013 Feet) * 1000

For Feet = 30,

Pound = (-18.016 + 1.013 * 30 ) * 1000 = 12374

Using R the 95% prediction interval is,

> newdata = data.frame(Feet = 30)

> predict.lm(m, newdata, interval = c("prediction"))
fit lwr upr
1 12.36945 2.635074 22.10382

The closest answer is,

• e. 8000-17000